Determine the new charge stored on the positive plate Please explain how to sovl

Question

Determine the new charge stored on the positive plate Please explain how to sovle the above problem:

Problem 26.23 A capacitor connected to a battery initially holds a charge of +q on its positive plate and -q on its negative plate. The electric field between the plates is initially E. A dielectric material is then inserted that polarizes in such a way as to produce an electric field of -0.30E (where the -sign indicates that this field opposes the initial field E). Part A Determine the new charge stored on the positive plate.

Explanation / Answer

E1=q/Ae0

q=E1Ae0

E=E1-E2=E1-0.3E=0.7 E1

new charge qn

E=qn/Ae0

qn=0.7E1Ae0=0.7q

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